What is the vertices f(x)=-2x^2+4x+1 ?

The vertices f(x)=-2x^2+4x+1 is Maximum (1,3)

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vertices f(x)=-2x^2+4x+1

vertices

f (x) = - 2 x^{2} + 4 x + 1

Maximum (1, 3)

Solution steps

y = - 2 x^{2} + 4 x + 1

The parabola parameters are:

a = - 2, b = 4, c = 1

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 ( - 2 )}

Simplify

- \frac{4}{2 ( - 2 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(1, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (1, 3)

3 x^{2} + 19 x - 40

(x + 4)^{2} = 12 y - 24

x^{2} - 10 x - 4 y + 21 = 0

f (x) = - 3 x^{2}

f (x) = - x^{2} + 4 x

What is the vertices f(x)=-2x^2+4x+1 ?
The vertices f(x)=-2x^2+4x+1 is Maximum (1,3)